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Unformatted text preview: 43. (a) and (b) Using Eq. 40-6 and the result of problem 3 in Chapter 39, we ﬁnd
∆E = Ephoton = hc
1240 eV · nm
= 2.55 eV .
486.1 nm Referring to Fig. 40-16, we see that this must be one of the Balmer series transitions (this fact
could also be found from Fig. 40-17). Therefore, nlow = 2, but what precisely is nhigh ?
= Elow + ∆E
+ 2.55 eV
22 which yields n = 4. Thus, the transition is from the n = 4 to the n = 2 state. ...
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